Question-164600

Question Number 164600 by mathlove last updated on 19/Jan/22

Answered by Zaynal last updated on 19/Jan/22

1. (1/5)arcsec((2/5)x)+C 2.((arctan((t^2 /8)))/(16)) + C ,C∈R 3.((arcsin(((7y)/3)))/7)+C,C∈R

$$\mathrm{1}.\:\frac{\mathrm{1}}{\mathrm{5}}\mathrm{arcsec}\left(\frac{\mathrm{2}}{\mathrm{5}}\boldsymbol{{x}}\right)+\boldsymbol{\mathrm{C}} \\ $$$$\mathrm{2}.\frac{\mathrm{a}\boldsymbol{{rctan}}\left(\frac{\boldsymbol{{t}}^{\mathrm{2}} }{\mathrm{8}}\right)}{\mathrm{16}}\:+\:\boldsymbol{{C}}\:,\boldsymbol{{C}}\in\mathbb{R} \\ $$$$\mathrm{3}.\frac{\boldsymbol{\mathrm{arcsin}}\left(\frac{\mathrm{7}\boldsymbol{\mathrm{y}}}{\mathrm{3}}\right)}{\mathrm{7}}+\boldsymbol{\mathrm{C}},\mathrm{C}\in\mathbb{R} \\ $$

Answered by Mathspace last updated on 20/Jan/22

1)I=∫ (dx/(x(√(4x^2 −25)))) ⇒I=_(2x=5t) (5/2) ∫ (dt/((5/2)t.5(√(t^2 −1)))) =∫ (dt/(t(√(t^2 −1))))=_(t=chu) ∫ ((shu)/(chu shu))du =∫ (du/(chu))=2∫ (du/(e^u +e^(−u) )) =_(e^u =y) 2∫ (dy/(y(y+y^(−1) )))=2∫(dy/(y^2 +1)) =2arctany +C =2arctan(e^u )+C u=argcht=ln(t+(√(t^2 −1))) ⇒ e^u =t+(√(t^2 −1))=((2x)/5)+(√((((2x)/5))^2 −1))⇒ I=2arctan(((2x)/5)+(√((((2x)/5))^2 −1)))+C

$$\left.\mathrm{1}\right){I}=\int\:\frac{{dx}}{{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{25}}} \\ $$$$\Rightarrow{I}=_{\mathrm{2}{x}=\mathrm{5}{t}} \:\:\frac{\mathrm{5}}{\mathrm{2}}\:\int\:\:\frac{{dt}}{\frac{\mathrm{5}}{\mathrm{2}}{t}.\mathrm{5}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\int\:\:\frac{{dt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}=_{{t}={chu}} \:\:\int\:\:\frac{{shu}}{{chu}\:{shu}}{du} \\ $$$$=\int\:\frac{{du}}{{chu}}=\mathrm{2}\int\:\frac{{du}}{{e}^{{u}} +{e}^{−{u}} } \\ $$$$=_{{e}^{{u}} ={y}} \:\:\mathrm{2}\int\:\:\frac{{dy}}{{y}\left({y}+{y}^{−\mathrm{1}} \right)}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\mathrm{2}{arctany}\:+{C} \\ $$$$=\mathrm{2}{arctan}\left({e}^{{u}} \right)+{C} \\ $$$${u}={argcht}={ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$${e}^{{u}} ={t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2}{x}}{\mathrm{5}}+\sqrt{\left(\frac{\mathrm{2}{x}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{1}}\Rightarrow \\ $$$${I}=\mathrm{2}{arctan}\left(\frac{\mathrm{2}{x}}{\mathrm{5}}+\sqrt{\left(\frac{\mathrm{2}{x}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{1}}\right)+{C} \\ $$

Answered by Eulerian last updated on 20/Jan/22

Solution: 1. ∫ ((1/x^2 )/( (√(4−((5/x))^2 )))) dx =^((5/x) = 2sin(𝛉)) −(2/5)∙∫ ((cos(θ))/( (√(4−4sin^2 (θ))))) dθ = −(2/5)∙∫ (1/( 2)) dθ = −(1/5)∙θ + C By substituting back: 1. −(1/5)∙sin^(−1) ((5/(2x))) + C

$$\: \\ $$$$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{1}.\:\int\:\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{4}−\left(\frac{\mathrm{5}}{\mathrm{x}}\right)^{\mathrm{2}} }}\:\mathrm{dx}\: \\ $$$$\: \\ $$$$\:\overset{\frac{\mathrm{5}}{\mathrm{x}}\:=\:\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)} {=}\:\:−\frac{\mathrm{2}}{\mathrm{5}}\centerdot\int\:\frac{\mathrm{cos}\left(\theta\right)}{\:\sqrt{\mathrm{4}−\mathrm{4sin}^{\mathrm{2}} \left(\theta\right)}}\:\mathrm{d}\theta \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:=\:−\frac{\mathrm{2}}{\mathrm{5}}\centerdot\int\:\frac{\mathrm{1}}{\:\mathrm{2}}\:\mathrm{d}\theta\: \\ $$$$\:\:\:\:\:\:\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\centerdot\theta\:+\:\mathrm{C} \\ $$$$\: \\ $$$$\:\mathrm{By}\:\mathrm{substituting}\:\mathrm{back}: \\ $$$$\:\mathrm{1}.\:−\frac{\mathrm{1}}{\mathrm{5}}\centerdot\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{2x}}\right)\:+\:\mathrm{C} \\ $$$$\: \\ $$$$\: \\ $$

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